The probability that at least one of A and B | vedclass

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(C) Given that the probability that at least one of $A$ and $B$ occurs is $P(A \cup B) = 0.6$. Also,the probability that $A$ and $B$ occur simultaneou

Vedclass · Accepted Answer

$1.1$

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(C) Given that the probability that at least one of $A$ and $B$ occurs is $P(A \cup B) = 0.6$. Also,the probability that $A$ and $B$ occur simultaneously is $P(A \cap B) = 0.3$. We know that $P(A \cup B) = 1 - P(A' \cap B')$,so $P(A' \cap B') = 1 - 0.6 = 0.4$. Using De Morgan's Law,$P(A' \cap B') = P((A \cup B)')$. We need to find $P(A') + P(B')$. Using the formula $P(A' \cup B') = P(A') + P(B') - P(A' \cap B')$,we have: $P(A' \cup B') = P((A \cap B)') = 1 - P(A \cap B) = 1 - 0.3 = 0.7$. Substituting the values into the formula: $0.7 = P(A') + P(B') - 0.4$. Therefore,$P(A') + P(B') = 0.7 + 0.4 = 1.1$.

The probability that at least one of $A$ and $B$ occurs is $0.6$. If $A$ and $B$ occur simultaneously with probability $0.3$,then $P(A') + P(B') = $

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